Now if we sub 3 into 1 to get A, and sub 2 into A to get B and then sub 4 into B to get C

[tex]C \rightarrow \lambda \gamma d = \lambda \gamma d[/tex]

Similarly if we, sub 2 into 3 to get A, and 1 into A to get B, and 4 into B to get C

[tex]C \rightarrow \lambda \gamma d = \lambda \gamma d[/tex]

I want to stop trying all the possible ways to get C now, because I want to look for a generalized way to show that they will all end up at the same point.

But more than this...what is the step of logic that connects the final equation C to proving the first two assumptions. I feel like this should prove the assumptions, but I don't know how exactly, or how exactly to express it.

With this second assumption, you are assuming the fact that you are supposed to prove! That won't get you anywhere. Instead, work with the first assumption, which is given: ##(a,b)## is a multiple of ##(c,d)##. So that means there is some constant ##\lambda## such that ##a = \lambda c## and ##b = \lambda d##.

By the way, note that the condition ##abcd \neq 0## means that all four of ##a,b,c,d## are nonzero, and therefore ##\lambda## is also nonzero.

Now we can rewrite the matrix as
$$\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} =
\begin{pmatrix}
\lambda c & \lambda d \\
c & d \\
\end{pmatrix}$$
From this, you can easily see that first column is a constant multiple of the second column. (What is the constant?)

With this second assumption, you are assuming the fact that you are supposed to prove! That won't get you anywhere. Instead, work with the first assumption, which is given: ##(a,b)## is a multiple of ##(c,d)##. So that means there is some constant ##\lambda## such that ##a = \lambda c## and ##b = \lambda d##.

By the way, note that the condition ##abcd \neq 0## means that all four of ##a,b,c,d## are nonzero, and therefore ##\lambda## is also nonzero.

Now we can rewrite the matrix as
$$\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} =
\begin{pmatrix}
\lambda c & \lambda d \\
c & d \\
\end{pmatrix}$$
From this, you can easily see that first column is a constant multiple of the second column. (What is the constant?)